3.240 \(\int \frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^{15/2}} \, dx\)

Optimal. Leaf size=148 \[ \frac {32 b^3 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{3003 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac {16 b^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{429 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{13 x^{13/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {12 b \tanh ^{-1}(\tanh (a+b x))^{7/2}}{143 x^{11/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

[Out]

32/3003*b^3*arctanh(tanh(b*x+a))^(7/2)/x^(7/2)/(b*x-arctanh(tanh(b*x+a)))^4+16/429*b^2*arctanh(tanh(b*x+a))^(7
/2)/x^(9/2)/(b*x-arctanh(tanh(b*x+a)))^3+12/143*b*arctanh(tanh(b*x+a))^(7/2)/x^(11/2)/(b*x-arctanh(tanh(b*x+a)
))^2+2/13*arctanh(tanh(b*x+a))^(7/2)/x^(13/2)/(b*x-arctanh(tanh(b*x+a)))

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Rubi [A]  time = 0.08, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2171, 2167} \[ \frac {16 b^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{429 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {32 b^3 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{3003 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac {2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{13 x^{13/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {12 b \tanh ^{-1}(\tanh (a+b x))^{7/2}}{143 x^{11/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^(5/2)/x^(15/2),x]

[Out]

(32*b^3*ArcTanh[Tanh[a + b*x]]^(7/2))/(3003*x^(7/2)*(b*x - ArcTanh[Tanh[a + b*x]])^4) + (16*b^2*ArcTanh[Tanh[a
+ b*x]]^(7/2))/(429*x^(9/2)*(b*x - ArcTanh[Tanh[a + b*x]])^3) + (12*b*ArcTanh[Tanh[a + b*x]]^(7/2))/(143*x^(1
1/2)*(b*x - ArcTanh[Tanh[a + b*x]])^2) + (2*ArcTanh[Tanh[a + b*x]]^(7/2))/(13*x^(13/2)*(b*x - ArcTanh[Tanh[a +
b*x]]))

Rule 2167

Int[(u_)^(m_)*(v_)^(n_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, -Simp[(u^(m + 1)*v^
(n + 1))/((m + 1)*(b*u - a*v)), x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m, n}, x] && PiecewiseLinearQ[u, v, x] && E
qQ[m + n + 2, 0] && NeQ[m, -1]

Rule 2171

Int[(u_)^(m_)*(v_)^(n_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, -Simp[(u^(m + 1)*v^
(n + 1))/((m + 1)*(b*u - a*v)), x] + Dist[(b*(m + n + 2))/((m + 1)*(b*u - a*v)), Int[u^(m + 1)*v^n, x], x] /;
NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && NeQ[m + n + 2, 0] && LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^{15/2}} \, dx &=\frac {2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{13 x^{13/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {(6 b) \int \frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^{13/2}} \, dx}{13 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ &=\frac {12 b \tanh ^{-1}(\tanh (a+b x))^{7/2}}{143 x^{11/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{13 x^{13/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {\left (24 b^2\right ) \int \frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^{11/2}} \, dx}{143 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}\\ &=\frac {16 b^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{429 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {12 b \tanh ^{-1}(\tanh (a+b x))^{7/2}}{143 x^{11/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{13 x^{13/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}+\frac {\left (16 b^3\right ) \int \frac {\tanh ^{-1}(\tanh (a+b x))^{5/2}}{x^{9/2}} \, dx}{429 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}\\ &=\frac {32 b^3 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{3003 x^{7/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^4}+\frac {16 b^2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{429 x^{9/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3}+\frac {12 b \tanh ^{-1}(\tanh (a+b x))^{7/2}}{143 x^{11/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}+\frac {2 \tanh ^{-1}(\tanh (a+b x))^{7/2}}{13 x^{13/2} \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 82, normalized size = 0.55 \[ \frac {2 \tanh ^{-1}(\tanh (a+b x))^{7/2} \left (-1001 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))+819 b x \tanh ^{-1}(\tanh (a+b x))^2-231 \tanh ^{-1}(\tanh (a+b x))^3+429 b^3 x^3\right )}{3003 x^{13/2} \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^4} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^(5/2)/x^(15/2),x]

[Out]

(2*ArcTanh[Tanh[a + b*x]]^(7/2)*(429*b^3*x^3 - 1001*b^2*x^2*ArcTanh[Tanh[a + b*x]] + 819*b*x*ArcTanh[Tanh[a +
b*x]]^2 - 231*ArcTanh[Tanh[a + b*x]]^3))/(3003*x^(13/2)*(-(b*x) + ArcTanh[Tanh[a + b*x]])^4)

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fricas [A]  time = 0.49, size = 78, normalized size = 0.53 \[ \frac {2 \, {\left (16 \, b^{6} x^{6} - 8 \, a b^{5} x^{5} + 6 \, a^{2} b^{4} x^{4} - 5 \, a^{3} b^{3} x^{3} - 371 \, a^{4} b^{2} x^{2} - 567 \, a^{5} b x - 231 \, a^{6}\right )} \sqrt {b x + a}}{3003 \, a^{4} x^{\frac {13}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2)/x^(15/2),x, algorithm="fricas")

[Out]

2/3003*(16*b^6*x^6 - 8*a*b^5*x^5 + 6*a^2*b^4*x^4 - 5*a^3*b^3*x^3 - 371*a^4*b^2*x^2 - 567*a^5*b*x - 231*a^6)*sq
rt(b*x + a)/(a^4*x^(13/2))

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giac [A]  time = 0.20, size = 97, normalized size = 0.66 \[ -\frac {\sqrt {2} {\left (\frac {429 \, \sqrt {2} b^{13}}{a} - 2 \, {\left (\frac {143 \, \sqrt {2} b^{13}}{a^{2}} + 4 \, {\left (\frac {2 \, \sqrt {2} {\left (b x + a\right )} b^{13}}{a^{4}} - \frac {13 \, \sqrt {2} b^{13}}{a^{3}}\right )} {\left (b x + a\right )}\right )} {\left (b x + a\right )}\right )} {\left (b x + a\right )}^{\frac {7}{2}} b}{3003 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {13}{2}} {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2)/x^(15/2),x, algorithm="giac")

[Out]

-1/3003*sqrt(2)*(429*sqrt(2)*b^13/a - 2*(143*sqrt(2)*b^13/a^2 + 4*(2*sqrt(2)*(b*x + a)*b^13/a^4 - 13*sqrt(2)*b
^13/a^3)*(b*x + a))*(b*x + a))*(b*x + a)^(7/2)*b/(((b*x + a)*b - a*b)^(13/2)*abs(b))

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maple [A]  time = 0.39, size = 151, normalized size = 1.02 \[ -\frac {2 \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{13 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {13}{2}}}-\frac {12 b \left (-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{11 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {11}{2}}}-\frac {4 b \left (-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{9 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {9}{2}}}+\frac {2 b \arctanh \left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{63 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {7}{2}}}\right )}{11 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )}\right )}{13 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^(5/2)/x^(15/2),x)

[Out]

-2/13/(arctanh(tanh(b*x+a))-b*x)/x^(13/2)*arctanh(tanh(b*x+a))^(7/2)-12/13*b/(arctanh(tanh(b*x+a))-b*x)*(-1/11
/(arctanh(tanh(b*x+a))-b*x)/x^(11/2)*arctanh(tanh(b*x+a))^(7/2)-4/11*b/(arctanh(tanh(b*x+a))-b*x)*(-1/9/(arcta
nh(tanh(b*x+a))-b*x)/x^(9/2)*arctanh(tanh(b*x+a))^(7/2)+2/63*b/(arctanh(tanh(b*x+a))-b*x)^2/x^(7/2)*arctanh(ta
nh(b*x+a))^(7/2)))

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maxima [A]  time = 0.42, size = 56, normalized size = 0.38 \[ \frac {2 \, {\left (16 \, b^{4} x^{4} - 40 \, a b^{3} x^{3} + 70 \, a^{2} b^{2} x^{2} - 105 \, a^{3} b x - 231 \, a^{4}\right )} {\left (b x + a\right )}^{\frac {5}{2}}}{3003 \, a^{4} x^{\frac {13}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^(5/2)/x^(15/2),x, algorithm="maxima")

[Out]

2/3003*(16*b^4*x^4 - 40*a*b^3*x^3 + 70*a^2*b^2*x^2 - 105*a^3*b*x - 231*a^4)*(b*x + a)^(5/2)/(a^4*x^(13/2))

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mupad [B]  time = 1.74, size = 413, normalized size = 2.79 \[ \frac {\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\frac {27\,b\,x\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{143}-\frac {106\,b^2\,x^2}{429}-\frac {{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{26}+\frac {20\,b^3\,x^3}{3003\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}+\frac {16\,b^4\,x^4}{1001\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}+\frac {128\,b^5\,x^5}{3003\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}+\frac {512\,b^6\,x^6}{3003\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^4}\right )}{x^{13/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^(5/2)/x^(15/2),x)

[Out]

((log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2*a)*exp(2*b*x) + 1))/2)^(1/2)*((27*b*
x*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x))/143 - (
106*b^2*x^2)/429 - (log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))
+ 2*b*x)^2/26 + (20*b^3*x^3)/(3003*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*e
xp(2*b*x) + 1)) + 2*b*x)) + (16*b^4*x^4)/(1001*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))
/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^2) + (128*b^5*x^5)/(3003*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2
*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^3) + (512*b^6*x^6)/(3003*(log(2/(exp(2*a)*exp(2*b*x) + 1))
- log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^4)))/x^(13/2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**(5/2)/x**(15/2),x)

[Out]

Timed out

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